4√5/3
2√6
√6
2√10/3
S.D=√Σ(x-x̄)²/n
x̄ = Σx/4 = (-1 + 0 + 1 + k)/4 = k/4Now √5 = √[((-1-k/4)² + (0-k/4)² + (1-k/4)² + (k-k/4)²)/4]→ 5 × 4 = 2(1+k²/16)² + 5k²/8→ 20 = 3k²/4→ k² = 24→ k = 2√6.